Camelot

Solution Approach

BFS for Knight Distances

Precompute the shortest knight distance between every pair of squares using BFS from each square. Let $\mathrm{kdist}[s][t]$ denote this distance. The board has at most $S = 30 \times 26 = 780$ squares.

King Distance

The king's distance is the Chebyshev distance: \[ \mathrm{kingDist}((r_k, c_k), (r, c)) = \max(|r_k - r|, |c_k - c|). \]

Enumeration

For each potential gathering square $g$, compute:

Option 1 -- King walks alone.

Option 2 -- A knight picks up the king.

Knight $i$ detours through pickup square $p$, where the king walks to $p$: \[ \mathrm{cost}_2 = \sum_{j \ne i} \mathrm{kdist}[\mathrm{knight}_j][g] + \mathrm{kdist}[\mathrm{knight}_i][p] + \mathrm{kdist}[p][g] + \mathrm{kingDist}(\mathrm{king}, p). \]

Take the minimum over all $g$, all knights $i$, and all pickup points $p$.

Pickup Point Search

To guarantee correctness, we search all squares as potential pickup points. Although this increases the time to $O(S^2 \cdot n)$, it remains feasible for the given constraints ($S \le 780$, $n \le 63$).

C++ Solution

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <climits>
using namespace std;

const int MAXR = 30, MAXC = 26;
int R, C;
int kdist[MAXR * MAXC][MAXR * MAXC];
int dx[] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dy[] = {-1, 1, -2, 2, -2, 2, -1, 1};

int idx(int r, int c) { return r * C + c; }

void bfs(int src) {
    queue<int> q;
    q.push(src);
    kdist[src][src] = 0;
    while (!q.empty()) {
        int u = q.front(); q.pop();
        int r = u / C, c = u % C;
        for (int d = 0; d < 8; d++) {
            int nr = r + dx[d], nc = c + dy[d];
            if (nr < 0 || nr >= R || nc < 0 || nc >= C) continue;
            int v = idx(nr, nc);
            if (kdist[src][v] == -1) {
                kdist[src][v] = kdist[src][u] + 1;
                q.push(v);
            }
        }
    }
}

int main() {
    scanf("%d %d", &R, &C);
    int total = R * C;

    // Read king position
    char col;
    int row;
    scanf(" %c %d", &col, &row);
    int kingR = row - 1, kingC = col - 'A';

    // Read knight positions
    int knights[65], numKnights = 0;
    while (scanf(" %c %d", &col, &row) == 2) {
        int kr = row - 1, kc = col - 'A';
        knights[numKnights++] = idx(kr, kc);
    }

    // BFS from every square
    memset(kdist, -1, sizeof(kdist));
    for (int i = 0; i < total; i++)
        bfs(i);

    // Replace unreachable entries with a large sentinel
    const int INF = 1000000;
    for (int i = 0; i < total; i++)
        for (int j = 0; j < total; j++)
            if (kdist[i][j] == -1)
                kdist[i][j] = INF;

    // Handle edge case: no knights
    if (numKnights == 0) {
        printf("0\n");
        return 0;
    }

    int ans = INT_MAX;

    // Try each gathering point
    for (int g = 0; g < total; g++) {
        int gr = g / C, gc = g % C;

        // Sum of all knight distances to g
        int base = 0;
        bool reachable = true;
        for (int i = 0; i < numKnights; i++) {
            if (kdist[knights[i]][g] >= INF) { reachable = false; break; }
            base += kdist[knights[i]][g];
        }
        if (!reachable) continue;

        // Option 1: king walks alone to g
        int kingAlone = base + max(abs(kingR - gr), abs(kingC - gc));
        ans = min(ans, kingAlone);

        // Option 2: one knight picks up king at some pickup point p
        for (int i = 0; i < numKnights; i++) {
            int saved = kdist[knights[i]][g];

            for (int p = 0; p < total; p++) {
                int pr = p / C, pc = p % C;
                int kingWalk = max(abs(kingR - pr), abs(kingC - pc));
                int knightDetour = kdist[knights[i]][p] + kdist[p][g];
                if (knightDetour >= INF) continue;
                int cost = base - saved + knightDetour + kingWalk;
                ans = min(ans, cost);
            }
        }
    }

    printf("%d\n", ans);
    return 0;
}

Complexity Analysis

• Time complexity:

  • BFS from every square: $O(S^2)$ where $S = R \times C \le 780$.

  • Main enumeration: for each gathering point ($O(S)$), for each knight ($O(n)$), for each pickup point ($O(S)$): $O(S^2 \cdot n)$.

  • Total: $O(S^2 \cdot n) \approx 780^2 \times 63 \approx 38{,}000{,}000$.

  • Space complexity: $O(S^2)$ for the knight distance table ($\approx 2.4$ MB).